Seating Four at a Round Table

Key idea. Fixing each person’s left and right neighbour fixes the clockwise cyclic order of the four diners, and nothing more. So a seating is a choice of four people together with a cyclic order of them, where rotations of the same order count as one.

Choosing the diners. There are $\binom{10}{4} = 210$ ways to choose who sits.

Arranging them. Fix one chosen person, say $A$, in some seat. This removes the rotational freedom. Reading clockwise from $A$, the other three can appear in any of $3! = 6$ orders.

Different orders give different seatings. Each order fixes, for every diner, who sits immediately to their right. Conversely, knowing each diner’s right-hand neighbour recovers the clockwise order starting from $A$.

Equivalently, the $4!$ assignments of the four people to four labelled seats fall into classes of $4$ rotations each. That gives $4!/4 = 3!$ seatings.

Note that left and right are distinguished. A seating and its mirror image therefore count as different: in the mirror image every person’s left and right neighbours are swapped.

Multiplying the two counts,

Generalisation. Seating $k$ of $n$ people at a round table, up to rotation, can be done in $\binom{n}{k}(k-1)! = \frac{n!}{k\,(n-k)!}$ ways.

If reflections were also identified, so that only the unordered pair of neighbours mattered, the count would be halved for $k\ge 3$. Here that gives $630$.